Problem: The prime factorization of 2160 is $2^4 \times 3^3 \times 5$. How many of its positive integer factors are perfect squares?
The prime factorization of a positive integer factor of 2160 is of the form $2^a\cdot3^b\cdot 5^c$ where $0\leq a\leq 4$, $0\leq b\leq 3$, and $0\leq c\leq 1$.  A positive integer is a perfect square if and only if all the exponents in its prime factorization are even.  Therefore, we are free to choose $a$ from the set $\{0,2,4\}$ and $b$ from the set $\{0,2\}$.  In total, we have $3\times 2=\boxed{6}$ choices for the exponents in the prime factorization of a perfect square factor of 2160.